Question: Define $f(x, y) = 4x - y$. Let $\vec{a} = (3, 1)$ and $\vec{v} = \left( 1, 1 \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
Our ultimate goal is to substitute $h = 0$ into the limit and get the answer. To do this, we first need to cancel out the $h$ in the denominator. Otherwise we're dividing by zero. How can we simplify the limit so that we get an $h$ on the top and an $h$ on the bottom, which we can then cancel? Let's plug in $\vec{a}$ and $\vec{v}$ to the limit. $ \lim_{h \to 0} \dfrac{f \left( (3, 1) + h \left( 1, 1 \right) \right) - f(3, 1)}{h}$ Now we can add together the vectors. $ \lim_{h \to 0} \dfrac{f \left( 3 + h, 1 + h \right) - f(3, 1)}{h}$ Let's evaluate $f$. $ \lim_{h \to 0} \dfrac{4(3 + h) - (1 + h) - (4(3) - 1)}{h}$ When we simplify, we can cancel out all the terms without an $h$. $ \lim_{h \to 0} \dfrac{4h - h}{h}$ Now we can cancel $h$ and calculate the limit. $\begin{aligned} \lim_{h \to 0} \dfrac{3h}{h} &= \lim_{h \to 0} 3 \\ \\ &= 3 \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = 3$.